JZPKIL

 题意描述：求 (sumlimits_{i=1}^n lcm(i,n)^ygcd(i,n)^x)

 数据范围：(1le n le 1e18,1le x,yle 3000)

大神题啊

先来一波反演

[egin{aligned} & sum_{i=1}^nlcm(i,n)^y gcd(i,n)^x\ &=sum_{i=1}^ngcd(i,n)^x (frac{in}{gcd(i,n)})^y\ &=sum_{i=1}^n gcd(i,n)^{x-y} i^y n^y\ &=n^ysum_{i=1}^n gcd(i,n)^{x-y} i^y\ &=n^ysum_{d|n}d^{x-y}sum_{i=1}^{frac{n}{d}}[gcd(i,frac{n}{d})==1](id)^y\ &=n^ysum_{d|n}d^xsum_{i=1}^{frac{n}{d}}[gcd(i,frac{n}{d})==1]i^y\ &=n^ysum_{d|n}d^xsum_{i=1}^{frac{n}{d}}i^ysum_{p|gcd(i,frac{n}{d})}mu(p)\ &=n^ysum_{d|n}d^xsum_{p|frac{n}{d}}mu(p) p^ysum_{i=1}^{frac{n}{dp}}i^y end{aligned}]

反演也就这些了，接下来代入伯努利数
有 $$sumlimits_{i=1}^n i^k=frac{1}{k+1}sumlimits_{i=1}^k inom{k+1}{i} B_i n^{k-i+1}$$
继续化柿子

[egin{aligned} & n^ysum_{d|n}d^xsum_{p|frac{n}{d}}mu(p) p^ysum_{i=1}^{frac{n}{dp}}i^y\ &=n^ysum_{d|n}d^xsum_{p|frac{n}{d}}mu(p) p^yfrac{1}{y+1}sum_{i=1}^yinom{y+1}{i}B_i(frac{n}{pd})^{y-i+1}\ &=frac{1}{y+1}sum_{i=1}^yinom{y+1}{i}B_i n^ysum_{d|n}d^xsum_{p|frac{n}{d}}mu(p) p^y (frac{n}{pd})^{y-i+1} end{aligned}]

令 (f(n,i)=n^ysumlimits_{d|n}d^xsumlimits_{p|frac{n}{d}}mu(p) p^y (frac{n}{dp})^{y-i+1})
发现后面这个 (f(n,i)) 对于每个 (n) 的质因子互不干扰，是个积性函数

现在只需要求出每个 (f(p_k^{e_k},i)) ，乘起来就行了

观察发现当且仅当 (p=1) 或 (p=p_k) 时柿子有意义

令 (N=p_k^{e_k})

那么 (f(N,i)=N^ysumlimits_{d|n}d^x((frac{N}{d})^{y-i+1}-p_k^{y}(frac{N}{p_kd})^{y-i+1}))

质因子分解需要 (Pollard)_(Rho)